Transcript 04-logic

Logic Gates
CS 202, Spring 2007
Epp, setions 1.4 and 1.5
Aaron Bloomfield
1
Review of Boolean algebra
• Just like Boolean logic
• Variables can only be 1 or 0
– Instead of true / false
2
Review of Boolean algebra
• Not_ is a horizontal bar above the number
–0
_ =1
– 1=0
• Or is a plus
–
–
–
–
0+0 = 0
0+1 = 1
1+0 = 1
1+1 = 1
• And is multiplication
–
–
–
–
0*0 = 0
0*1 = 0
1*0 = 0
1*1 = 1
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Review of Boolean algebra
___
• Example: translate (x+y+z)(xyz) to a Boolean
logic expression
– (xyz)(xyz)
• We can define a Boolean function:
– F(x,y) = (xy)(xy)
• And then write a “truth table” for it:
x
y
F(x,y)
1
1
0
1
0
0
0
1
0
0
0
0
4
Basic logic gates
• Not
x
• And
x
y
xy
• Or
x
y
x+y
• Nand
x
y
xy
• Nor
x
y
x+y
• Xor
x
y
xÅy
x
x
y
z
x
y
z
xyz
x+y+z
5
Converting between circuits and
equations
• Find the output of the following circuit
x
y
x+y
y
y
(x+y)y
__
• Answer: (x+y)y
– Or (xy)y
6
Converting between circuits and
equations
• Find the output of the following circuit
x
x
y
y
xy
xy
___
__
• Answer: xy
– Or (xy) ≡ xy
7
Converting between circuits and
equations
•
Write the circuits for the following
Boolean algebraic expressions
__
a) x+y
x
y
x
x+y
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Converting between circuits and
equations
•
Write the circuits for the following
Boolean
algebraic
expressions
_______
b) (x+y)x
x
y
x+y
x+y
(x+y)x
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Writing xor using and/or/not
• p Å q  (p  q)  ¬(p  q)
____
• x Å y  (x + y)(xy)
x
y
x+y
xy
x
y
xÅy
1
1
0
1
0
1
0
1
1
0
0
0
(x+y)(xy)
xy
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Converting decimal numbers to
binary
• 53 = 32 + 16 + 4 + 1
= 25 + 24 + 22 + 20
= 1*25 + 1*24 + 0*23 + 1*22 + 0*21 + 1*20
= 110101 in binary
= 00110101 as a full byte in binary
• 211= 128 + 64 + 16 + 2 + 1
= 27 + 26 + 24 + 21 + 20
= 1*27 + 1*26 + 0*25 + 1*24 + 0*23 + 0*22 +
1*21 + 1*20
= 11010011 in binary
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Converting binary numbers to
decimal
• What is 10011010 in decimal?
10011010
= 1*27 + 0*26 + 0*25 + 1*24 + 1*23 +
0*22 + 1*21 + 0*20
= 27 + 24 + 23 + 21
= 128 + 16 + 8 + 2
= 154
• What is 00101001 in decimal?
00101001 = 0*27 + 0*26 + 1*25 + 0*24 + 1*23 +
0*22 + 0*21 + 1*20
= 25 + 23 + 20
= 32 + 8 + 1
= 41
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A note on binary numbers
• In this slide set we are only dealing with
non-negative numbers
• The book (section 1.5) talks about two’scomplement binary numbers
– Positive (and zero) two’s-complement binary
numbers is what was presented here
– We won’t be getting into negative two’scomplmeent numbers
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How to add binary numbers
• Consider adding two 1-bit binary numbers x and y
–
–
–
–
0+0 = 0
0+1 = 1
1+0 = 1
1+1 = 10
x
0
0
1
1
y
0
1
0
1
Carry Sum
0
0
0
1
0
1
1
0
• Carry is x AND y
• Sum is x XOR y
• The circuit to compute this is called a half-adder
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The half-adder
• Sum = x XOR y
• Carry = x AND y
x
y
x
y
x
y
0
0
0
0
0
1
0
1
1
0
0
1
1
1
1
0
Sum
Carry
Carry Sum
Sum
Carry
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Using half adders
• We can then use a half-adder to compute
the sum of two Boolean numbers
1
1
+1
?
0
1
1
0
0
0 0
1 0
1 0
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How to fix this
• We need to create an adder that can take a
carry bit as an additional input
– Inputs: x, y, carry in
– Outputs: sum, carry out
• This is called a full adder
– Will add x and y with a half-adder
– Will add the sum of that to the
carry in
• What about the carry out?
– It’s 1 if either (or both):
– x+y = 10
– x+y = 01 and carry in = 1
x y c carry sum
1 1 1 1
1
1 1 0 1
0
1
1
0
0
0
0
1
1
1
0
1
0
1
0
1
0
0
1
0
1
0 0 1
0 0 0
0
0
1
0
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The full adder
• The “HA” boxes are
half-adders
c
x
y
X
s1
X
Y
HA
Y
HA
S
S
x
1
1
1
1
0
0
0
0
y
1
1
0
0
1
1
0
0
c
1
0
1
0
1
0
1
0
s1
0
0
1
1
1
1
0
0
c1 carry sum
1
1
1
1
1
0
0
1
0
0
0
1
0
1
0
0
0
1
0
0
1
0
0
0
s
C
C
S
C
c1
c
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The full adder
• The full circuitry of the full adder
c
s
x
y
c
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Adding bigger binary numbers
• Just chain full adders together
Y
HA
s0
S
C
C
FA
s1
S
X
Y
C
C
FA
s2
S
X
Y
C
C
FA
S
X
Y
...
x0
y0
x1
y1
x2
y2
x3
y3
X
C
s3
c
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Adding bigger binary numbers
• A half adder has 4 logic gates
• A full adder has two half adders plus a OR gate
– Total of 9 logic gates
• To add n bit binary numbers, you need 1 HA and
n-1 FAs
• To add 32 bit binary numbers, you need 1 HA
and 31 FAs
– Total of 4+9*31 = 283 logic gates
• To add 64 bit binary numbers, you need 1 HA
and 63 FAs
– Total of 4+9*63 = 571 logic gates
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More about logic gates
• To implement a logic gate in hardware,
you use a transistor
• Transistors are all enclosed in an “IC”, or
integrated circuit
• The current Intel Pentium IV processors
have 55 million transistors!
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Flip-flops
• Consider the following circuit:
• What does it do?
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Memory
• A flip-flop holds a single bit of memory
– The bit “flip-flops” between the two NAND
gates
• In reality, flip-flops are a bit more
complicated
– Have 5 (or so) logic gates (transistors) per flipflop
• Consider a 1 Gb memory chip
– 1 Gb = 8,589,934,592 bits of memory
– That’s about 43 million transistors!
• In reality, those transistors are split into 9
ICs of about 5 million transistors each
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Hexadecimal
• A numerical
from 0-15
range
– Where A is 10, B is 11,
… and F is 15
• Often written with a
‘0x’ prefix
• So 0x10 is 10 hex, or
16
– 0x100 is 100 hex, or
256
• Binary numbers easily
translate:
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From
ThinkGeek
(http://www.thinkgeek.com)
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Also from
ThinkGeek
(http://www.thinkgeek.com)
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DEADBEEF
• Many IBM machines would fill allocated
(but uninitialized) memory with the hexadecimal pattern 0xDEADBEEF
– Decimal -21524111
– See http://www.jargon.net/jargonfile/d/DEADBEEF.html
• Makes it easier to spot in a debugger
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Also also from ThinkGeek
(http://www.thinkgeek.com)
• 0xDEAD = 57005
• Now add one to that...
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